Asked by Annale
Find the indefinite integral.
∫ x^2(1-x)^6 dx
I am so lost on this. I let u=1-x so du=-dx.
After that, I'm not sure what to do.
∫ x^2(1-x)^6 dx
I am so lost on this. I let u=1-x so du=-dx.
After that, I'm not sure what to do.
Answers
Answered by
Damon
go to
http://www.wolframalpha.com/input/?i=integrate+x%2F%281-x%29
type in:
integrate x^2(1-x)^6
It is a serious mess !!!
http://www.wolframalpha.com/input/?i=integrate+x%2F%281-x%29
type in:
integrate x^2(1-x)^6
It is a serious mess !!!
Answered by
Damon
The only thing I know to do is multiply (1-x)^6 out (use binary coefficients)
then you have a whole string of simple ones
which is why the answer at Wolfram starts with someting times x^9
then you have a whole string of simple ones
which is why the answer at Wolfram starts with someting times x^9
Answered by
Damon
(1-x)^6
= 1 - 6x + 15x^2 - 20x^3 etc
multiply by x^2
so
x^2 - 6x^3 + 15x^4-20x^5 .....etc
THEN integrate
(1/3)x^3 - (3/2)x^4 + 4x^5 -(10/3)x^6 ... etc
= 1 - 6x + 15x^2 - 20x^3 etc
multiply by x^2
so
x^2 - 6x^3 + 15x^4-20x^5 .....etc
THEN integrate
(1/3)x^3 - (3/2)x^4 + 4x^5 -(10/3)x^6 ... etc
Answered by
Damon
whoop, that is 3 x^5 not 4 x^5
Answered by
Damon
Otherwise I am agreeing with Wolfram
Answered by
Damon
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